### Video Transcript

Use the integral test to determine
whether the series, which is the sum from π equals one to β of one over π,
converges or diverges.

Letβs start by recalling the
integral test. The integral test tells us that if
π is a continuous, positive, decreasing function on the interval between π and β
and that π of π is equal to π π. Then, if the integral from π to β
of π of π₯ with respect to π₯ is convergent, so is the sum from π equals π to β
of π π. And secondly, if the integral from
π to β of π of π₯ with respect to π₯ is divergent, so is the sum from π equals π
to β of π π. Now, in our case, our series is the
sum from π equals one to β of one over π. Therefore, π π is equal to one
over π. Using the fact that π of π is
equal to π π, we can say that π of π₯ is equal to one over π₯.

Now, we need to check that π of π₯
is a continuous, positive, and decreasing function on the interval between π and
β. Noting that since our sum goes from
π equals one, π must be equal to one. Letβs start by checking the
continuity of our function over this interval. The only discontinuity of π of π₯
occurs when π₯ is equal to zero. However, zero is not in our
interval between one and β. Therefore, this function must be
continuous between one and β. On this interval, π₯ is always
positive. Therefore, one over π₯ must also be
positive. And so, weβve satisfied that
condition too. Now, if π₯ starts at one and gets
larger and larger and larger, then one over π₯ will get smaller and smaller and
smaller. Therefore, we can say that this is
a decreasing function on our interval.

Now that weβve satisfied these
three conditions, weβre able to use the integral test. We need to work out whether the
integral from one to β of one over π₯ with respect to π₯ is convergent or
divergent. Now, we know that the differential
of the natural logarithm of π₯ with respect to π₯ is equal to one over π₯. Therefore, the natural logarithm of
π₯ is the antiderivative of one over π₯. Hence, we have that our integral is
equal to the natural logarithm of π₯ between one and β. When using the bound of β, we need
to take the limit as π₯ tends to β of the natural logarithm of π₯. Then, for our lower bound of one,
we simply subtract the natural logarithm of one.

Now, letβs evaluate this limit, the
natural logarithm as an increasing function. Therefore, as π₯ gets larger, the
natural logarithm of π₯ also gets larger. Hence, we can say that this limit
is also equal to β. We also have that the natural
logarithm of one is equal to zero. However, regardless of what
constant this term is, this will not affect our answer of β. This tells us that our integral
must be divergent. Hence, by the integral test, we can
say that the sum from π equals one to β of one over π is divergent.